IPMAT Indore 2022 · Quantitative Ability · Question 21

Step 1 — recall what divisibility by 6 requires. A number is divisible by 6 exactly when it is divisible by both 2 and 3:

by 3: the digit sum is a multiple of 3 — this depends only on which digits appear, not their order;

by 2: the last digit is even — this depends entirely on position.

Step 2 — see what rearranging preserves. Any rearrangement of a,b,c,d,e keeps the digit sum, so divisibility by 3 always survives. Only the last-digit condition can break.

Given \overline{abcde} is divisible by 6, we know e is even and a+b+c+d+e\equiv 0 (\mod 3).

Step 3 — test each option’s last digit.

(a) \overline{edcba} — ends in a. Nothing forces a to be even. ✘ This is the answer.

(b) \overline{eee} — ends in e, which is even ✔; digit sum 3e is a multiple of 3 ✔. Always divisible by 6.

(c) \overline{bbadcacede} — ends in e, even ✔; its digit sum is 2\left( a+b+c+d+e \right)+\left( c+e \right)… more simply, it ends in e and we can check the sum is a multiple of 3: the ten digits are b,b,a,d,c,a,c,e,d,e, i.e. 2\left( a+b+c+d+e \right), which is a multiple of 3 ✔.

(d) \overline{cdbae} — ends in e, even ✔; same digits, same sum ✔.

Concrete counterexample for (a). Take \overline{abcde}=12354: digit sum 15 ✔ and last digit 4 is even ✔, so it is divisible by 6. Reversed, \overline{edcba}=45321, which is odd — not divisible by 2, hence not by 6.

Answer: Option 1 — \overline{edcba}

Shortcut. Only the last digit can go wrong. Every option is a rearrangement of the same digits, so the digit sum — and with it divisibility by 3 — is automatic. Scan the options for the one that does not end in e:

Answer (a), found in about five seconds. The transferable idea: when a question rearranges digits, split the divisibility rule into a position-free part (digit sum) and a position-dependent part (final digits), and test only the second.